Given a diprotic acid, H, A, with two ionization constants of K1 = 49 x 10-...

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Answer 1

Answer #1

Ka1 = 4.9 x 10^-4

pKa1 = -log Ka1 = 3.31

pKa2 = - log Ka2 = 11.58

NaHA is middle salt . so

pH = pKa1 + pKa / 2

= (3.31 + 11.58) / 2

pH = 7.45

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Answer 2

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